IMO Practice Test: Continuity and Differentiability | Class 12 Mathematics

Question 1 of 6 hard

If $f(x)=\begin{cases}\frac{\sin(a+1)x+\sin x}{x} & x<0 \\ c & x=0 \\ \frac{\sqrt{x+bx^2}-\sqrt{x}}{bx^{3/2}} & x>0\end{cases}$ is continuous at $x=0$, then $(a,b,c)=$

$(-3/2,\text{ any non-zero},1/2)$

$(0,0,0)$

$(1,1,1)$

$(-1/2,0,3/2)$

Explanation: LHL: as $x\to0^-$: $\frac{\sin(a+1)x+\sin x}{x}\to(a+1)+1=a+2$. RHL: $\frac{\sqrt{x+bx^2}-\sqrt x}{bx^{3/2}}=\frac{\sqrt x(\sqrt{1+bx}-1)}{bx^{3/2}}\approx\frac{bx/2}{bx}=1/2$. Continuity: $a+2=c=1/2\Rightarrow a=-3/2,c=1/2$, $b$ can be any non-zero value.

Question 2 of 6 hard

Number of points where $f(x)=\max(\sin x, \cos x)$ is not differentiable on $[0,2\pi]$:

2

3

4

1

Explanation: $\sin x=\cos x$ at $x=\pi/4$ and $x=5\pi/4$. At these points, the max function has a corner — not differentiable. 2 points.

Question 3 of 6 hard

If $f(x+y)=f(x)f(y)$ for all $x,y$, and $f'(0)=2$, $f(3)=4$, then $f'(3)=$

8

2

4

6

Explanation: From functional equation, $f(x)=f(0)^x$ and differentiating: $f'(x)=f(x)f'(0)=2f(x)$. $f'(3)=2f(3)=2\times4=8$.

Question 4 of 6 hard

For $y=\sqrt{\sin x+\sqrt{\sin x+\sqrt{\sin x+...}}}$, $\frac{dy}{dx}=$

$\frac{\cos x}{2y-1}$

$\frac{\cos x}{2y}$

$\frac{\sin x}{2y-1}$

$\cos x$

Explanation: $y=\sqrt{\sin x+y}$, so $y^2=\sin x+y$. Differentiate: $2y\,dy/dx=\cos x+dy/dx$. $dy/dx(2y-1)=\cos x$. $dy/dx=\frac{\cos x}{2y-1}$.

Question 5 of 6 hard

If $y=\tan^{-1}\frac{\sqrt{1+x^2}-1}{x}$, then $\frac{dy}{dx}=$

$\frac{1}{2(1+x^2)}$

$\frac{1}{1+x^2}$

$\frac{2}{1+x^2}$

$\frac{1}{\sqrt{1+x^2}}$

Explanation: Put $x=\tan\theta$: $\frac{\sec\theta-1}{\tan\theta}=\frac{1-\cos\theta}{\sin\theta}=\tan(\theta/2)$. So $y=\theta/2=\frac{1}{2}\tan^{-1}x$. $dy/dx=\frac{1}{2(1+x^2)}$.

Question 6 of 6 hard

The function $f(x)=x|x|$ is:

Differentiable everywhere and $f'(0)=0$

Differentiable everywhere and $f'(0)=1$

Not differentiable at $x=0$

Continuous but not differentiable at $x=0$

Explanation: For $x>0$: $f(x)=x^2$, $f'(x)=2x$. For $x<0$: $f(x)=-x^2$, $f'(x)=-2x$. At $x=0$: LHD$=0$=RHD. So $f'(0)=0$.

IMO Practice Test — Continuity and Differentiability